Marshall Hall’s conjecture

Conjecture. (Marshall Hall, Jr.). With the exception of $n^2$ being a perfect sixth power, for any positive integer $n$, the inequality $|n^2-m^3|>C\sqrt{m}$, (with $m$ also being a positive integer and $C$ being a number less than 1 that nears 1 as $n$ tends to infinity) always holds.

The reason for the exception of perfect sixth powers (those cases of $n$ for which there is a solution to $n^2=h^6$ in integers) is a simple consequence of associativity: if $n^2=h^6$, then $h^6=h^2{h}^2{h}^2=h^3{h}^3$. Then $m=h$ and $n^2-m^3=0$. For example, $8^2-4^3=0$.

For small $n$, $C$ can’t be exactly 1. For example, $3^2-2^3=1$, and $\sqrt{2}>1$. But even among the smaller numbers, the conjecture generally holds even with $C=1$. After $n=3$, the next counterexample (that is not a perfect sixth power) to $C=1$ is $n=378661$, with the corresponding $m=5234$ producing a difference of just 17. A078933 in Sloane’s OEIS lists smaller values of $m$ with cubes being at a distance from the nearest square that is less than $\sqrt{m}$. Noam Elkies has found some fairly large counterexamples to setting $C=1$, such as $n=447884928428402042307918$ and $m=5853886516781223$, the difference between the square of the former and the cube of the latter being a relatively small 1641843.